convex duality in math finance - homepages at wmuhomepages.wmich.edu/~zhu/mf/lecture1a.pdf ·...

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IntroductionConstrained optimization problem

SubdifferentialLagrange multiplier theoremConvex sets and functions

CONVEX DUALITY IN MATH FINANCE1. Constrained Optimization and Lagrange Multipliers

Peter Carr and Qiji Zhu

Morgen Stanley/NYU andWestern Michigan University

September 2, 2015

Peter Carr and Qiji Zhu CONVEX DUALITY IN MATH FINANCE

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GoalVariational approachThe meaning of Lagrange MultipliersDualitySupport vs tangentNotation

Goal

• Increasing concave utilities and convex risk measures arecommon in financial problems.

• Moreover, no-arbitrage often implies the price of manyfinancial derivatives are convex in their parameters.

• Hence convex analysis methods are intrinsically involved inmany financial problems.

• We intend to provide a perspective of many important issuesin mathematical finance based on convex duality theory.


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Lagrange multiplier rules (LMR)

We start with Lagrange multiplier rules because

• Many financial problems can naturally be formulated asconstrained optimization problems.

• Lagrange multipliers are convenient tools for solving thoseproblems.

• They also bridge those financial problems to convex duality.


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Variational approach

For financial applications it is convenient to take the variationalanalysis view of the Lagrange multiplier. Define

v(y) = inff(x) : g(x) = y

and assume x0 is a solution to v(0).Then

x0 ∈ argmin[f(x)− v(g(x))]

Assuming all involved are smooth then

f ′(x0)− v′(g(x0))g′(x0) = 0

revealing λ = −v′(g(x0)) as a Lagrange multiplier.


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Why variational approach

• Explains the Lagrange multipliers as shadow prices.

• Leads naturally to duality.

• Reveals that in dealing with minimization problems what’sessential is a lower horizontal support rather than tangent.


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Economic meaning

D. Gale, gives the following economic explanation and provided arigorous proof for convex problems in 1967.

• View y as constrains in resources and −f as output in aneconomy.

• The Lagrange multiplier λ = −v′(g(x0)) then reflects themarginal gain of the output function with respect to theresource constraints.

• Following this observation, if we penalize the resourceutilization with the Lagrange multiplier (shadow price) thenthe constrained optimization problem can be converted to anunconstrained one.


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Dual problem

For simplicity consider the linear form of the above economicoutput problem with resource constraint:

maxx

⟨c, x⟩ s.t. Ax = b, x ≥ 0.

Consider the flip side of the problem: what is the fair price(vector) p to buy the resources. Seller is willing to sell only whenA⊤p ≥ c. So we get the dual problem

minp

⟨p, b⟩ s.t. A⊤p ≥ c.


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Dual problem

Clearly for any feasible pair (x, p) for the primal-dual problem wehave weak duality

⟨p, b⟩ = ⟨p,Ax⟩ = ⟨A⊤p, x⟩ ≥ ⟨c, x⟩.

If equality holds at (x, p) then it is easy to check that x solves theprimal and p solves the dual and is a Lagrange multiplier of theprimal.


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Support vs tangent

• In general, v is neither convex nor smooth so we cannot usethe usual Fermat’s rule to get necessary optimality conditions.

• But this also get people to think and to realize a horizontalsupport from below is what really needed.

• This is one of the most important observation that leads tomuch of the modern development of non-smooth andvariational analysis.

• Related framework (e.g. subdifferential, conjugate) andtechniques (of handling them) need to be developed though.


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Spaces

• Let X be finite dimensional complete normed vector spaces –Banach spaces.(Examples: RN , RV (Ω,F , P )).

• We denote Br(x) := y ∈ X | ∥y − x∥ ≤ r the closed ballaround x with radius r.

• We use X∗ to denote the dual of X which is also a Banachspace.

• The pairing between x∗ ∈ X∗ and x ∈ X is denoted by⟨x∗, x⟩.

• Let ≤K be the partial order induced by a closed cone K ⊂ X.

• The polar cone of K is defined byK+ := x∗ ∈ X∗ : ⟨x∗, x⟩ ≥ 0, ∀x ∈ K


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Functions

We consider extended valued functionf : X 7→ R ∪ +∞ andoften assume lower semicontinuous condition (lsc).

Lower semicontinuity

We say f is lsc at x ∈ X if

lim infy→x

f(y) ≥ f(x).

We say f is lsc if it is lsc at every point in X.


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Functions

A convenient characterization of lsc is

Characterization of lower semicontinuity

An extended valued function f : X 7→ R ∪ +∞ is lsc iff itsepigraph

epi f := (x, r) ∈ X × R | f(x) ≤ r

is closed.

We will often explore the relationship between functions and theirrelated sets and vice versa.


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Sup functions

Lower semicontinuity of the sup

If fα : X 7→ R ∪ +∞ is a family of lsc functions then so issupα fα.

Proof:epi sup

αfα =

∩α

epi fα.


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lsc functions related to sets

Let S ⊂ X be closed. Then, all the following functions are lsc• Indicator function:

ιS(x) =

0 x ∈ S

+∞ x ∈ S.

• The negative of the characteristic function:

χS(x) =

1 x ∈ S

0 x ∈ S.

• Distance function:

dS(x) = inf∥x− y∥ : y ∈ S


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Mappings

Similar concept also extend to mapping f : X 7→ Y when theimage space Y has the partial order ≤K generated by a closedconvex cone K ⊂ Y .

Lower semicontinuity

We say f is lsc at x ∈ X if

epi f := (x, y) ∈ X × Y | f(x) ≤K y

is closed.


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Constrained optimization problemLagrange multipliers

Constrained optimization problem

Let X,Y and Z be finite dimensional Banach spaces and let ≤K

be a partial order in Y induced by a closed convex cone K ⊂ Y .Consider constrained optimization problem

v(y, z) = inff(x) : g(x) ≤K y, h(x) = z, x ∈ C, (1)

where y ∈ Y , z ∈ Z, f, g are lsc, h is continuous and C ⊂ X isclosed. Denote S(y, z) the (possibly empty) solution set of (1).We are interested in (y, z) = (0, 0) but need to imbed the problemin a larger class of problems.


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Constrained optimization problemLagrange multipliers

Lagrange multiplier

Lagrange multiplier

We say that λ is a Lagrange multiplier for problem v(0, 0) if(i) (nonnegativity) λ ∈ K+ × Z∗ and,(ii) (unconstrained optimum)

f(x) + ⟨λ, (g(x), h(x))⟩ ≥ v(0, 0).

We denote the set of Lagrange multipliers for problem v(0, 0) by

Λ(0, 0).


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SubdifferentialGeometryExamples

Subdifferential

As noted above, v may not be differentiable. Subdifferential is asubstitute for the derivative.

Subdifferential

The subdifferential of a lower semi-continuous function ϕ atx ∈ dom ϕ is defined by

∂ϕ(x) = x∗ ∈ X∗ : ϕ(y)− ϕ(x) ≥ ⟨x∗, y − x⟩.

Subdifferential was initially defined for convex functions but worksfor general function as well.


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Geometry

Derivative has two main usage

1. Derivative act as a linear approximation.

2. In extreme problems, derivative =0 signals horizontal supportfrom below or cap from above.

The idea here is for minimization problem, support from belowonly is enough which leads to subdifferential and does not requiresmoothness.


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Examples

• If f ∈ C1(X) then ∂f(x) = f ′(x).• ∂∥ · ∥(0) = B1(0) = [−1, 1].

• ∂(·)+(0) = [0, 1].

• ∂(·)−(0) = [−1, 0].


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Lagrange multiplier theoremProofComplementary slacknessExample: Separation theoremExample: Sandwich theorem

Lagrange multiplier theorem (LMT)

Characterization of Lagrange multiplier

Λ(0, 0) = −∂v(0, 0).


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Part 1: Λ(0, 0) ⊇ −∂v(0, 0)

• Suppose that λ ∈ −∂v(0, 0).

• y → v(y, 0) is non-increasing w.r.t. ≤K .

• Thus, for any y ∈ K,

0 ≥ v(y, 0)− v(0, 0) ≥ ⟨−λ, (y, 0)⟩

so that λ ∈ K+ × Z∗.(Property (i))

• Property (ii) follows from, for all x ∈ C,

f(x) + ⟨λ, (g(x), h(x))⟩≥ v(g(x), h(x)) + ⟨λ, (g(x), h(x))⟩ ≥ v(0, 0)


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Part 2: Λ(0, 0) ⊆ −∂v(0, 0)

Suppose λ satisfies conditions (i) and (ii). Then we have, for anyx ∈ C, g(x) ≤K y and h(x) = z,

f(x) + ⟨λ, (y, z)⟩ ≥ f(x) + ⟨λ, (g(x), h(x))⟩ ≥ v(0, 0).

Taking the infimum with constraints x ∈ C, g(x) ≤K y andh(x) = z, we arrive at

v(y, z) + ⟨λ, (y, z)⟩ ≥ v(0, 0).

Therefore, −λ ∈ ∂v(0, 0).


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When S(0, 0) = ∅

Lagrange multiplier theorem with complementary slackness

−λ ∈ ∂v(0, 0) and x0 ∈ S(0, 0) if and only if

(i) (nonnegativity) λ ∈ K+ × Z∗;

(ii) (complementary slackness) ⟨λ, (g(x0), h(x0))⟩ = 0;

(iii) (unconstrained optimum) function

x 7→ f(x) + ⟨λ, (g(x), h(x))⟩

attains minimum at x0.


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Proof

We have already seen that −λ ∈ ∂v(0, 0) is characterized by

f(x) + ⟨λ, (g(x), h(x))⟩ ≥ v(0, 0) (*)

When x0 ∈ S(0, 0), use v(0, 0) = v(g(x0), h(x0)) = f(x0) we get(ii) and (iii). Conversely (ii) and (iii) with (*) clearly implies thatx0 ∈ S(0, 0).


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Need to look for a local version

If the Lagrange multiplier - as defined before - exists, will result ina global unconstrained problem.

• In general, this is not to be expected and a local version ofthe subdifferential is often important.

• Financial problems are often convex and the above definitionnaturally fits.


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Hahn-Banach separating hyperplane theorem

Hahn-Banach separating hyperplane theorem

Let X be a Banach space and let C1 and C2 be convex subsets ofX. Suppose that

C2 ∩ intC1 = ∅. (2)

Then there exists λ ∈ X∗\0 such that, for all x ∈ C1 andy ∈ C2,

⟨λ, y⟩ ≥ ⟨λ, x⟩. (3)


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Proof

In examples below we always assume the existence of LM.WOLG assume 0 ∈ intC1. Then the gauge function of C1,

γC1(x) := inft | x ∈ tC1

has the property that γC1(x) < 1 iff x ∈ int C1 anddom γC1 = X.Clearly

v(0) = minx,y

f(x) | h(x, y) = 0, y ∈ cl C2 ≥ 0, (4)

where f(x) = γC1(x)− 1, h(x, y) = x− y.


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Proof

Let λ ∈ X∗ be a Lagrange multiplier. Then, for all x ∈ X andy ∈ clC2,

γC1(x)− 1 + ⟨λ, y − x⟩ ≥ v(0) ≥ 0. (5)

or

⟨λ, y⟩ ≥ ⟨λ, x⟩+ 1− γC1(x). (6)

Letting x = 0 we see that λ = 0. Noting that x ∈ C1 implies that1− γC1(x) ≥ 0 we derive the separation theorem.


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• The Lagrange multiplier plays the role of the separatinghyperplane.

• Existence of an optimal solution is neither expected norneeded.

• Derivative information of the function involved is not needed.


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Sandwich theorem

Sandwich theorem

Let X and Y be Banach spaces, let f and g be convex lscfunctions, and let A : X → Y be a linear mapping. Suppose thatf ≥ −g A and (CQ) 0 ∈ int(dom g −Adom f). Then thereexists an affine function α : X → R of the form

α(x) = ⟨A∗y∗, x⟩+ c

satisfyingf ≥ α ≥ −g A.


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Sketch of Proof

Consider minimization problem

v(y, z) = minf(x) + g(Ax+ y)− z (7)

= minf(x) + r : u−Ax = y, g(u)− r ≤ z.

f ≥ −g A implies that v(0, 0) ≥ 0.CQ implies a Lagrange multiplier of the form (y∗, µ) ∈ Y ∗ ×R+

exists.


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Sketch of Proof

By the definition of the multiplier, for x ∈ X, u ∈ Y and r ≥ g(u),

f(x) + r + ⟨y∗, u−Ax⟩+ µ(g(u)− r) ≥ v(0, 0) ≥ 0. (8)

Letting u = Ax′ and r = g(Ax′) in (8) we have, for x, x′ ∈ X,

f(x)− ⟨A∗y∗, x⟩ ≥ −g(Ax′)− ⟨A∗y∗, x′⟩.

Thus,

a := infx[f(x)− ⟨A∗y∗, x⟩] ≥ b := sup

x′[−g(Ax′)− ⟨A∗y∗, x′⟩].

Picking any c ∈ [a, b], the affine function α(x) = ⟨A∗y∗, x⟩+ cseparates f and −g A.


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• The Lagrange multiplier induces the separating affine function.

• If A = I, g = ιcl C2 , f = γC1 − 1 we recover the separationtheorem.

• Thus, sandwich theorem provide more information (usefullater in FTAP).


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DefinitionsCharacterizationsOperations perserving convexityOptimal value functionSome useful convex functionsConvexification

Enter convex analysis

• We have seen that the existence of LM is equivalent to

∂v(0, 0) = ∅.

• In general this is hard to verify.

• However, in the class of convex functions this is relatively easy.

Let’s get into the details.


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Convex sets

Convex sets

We say C ⊂ X is convex if for any x, y ∈ C and λ ∈ [0, 1],

λx+ (1− λ)y ∈ C.

Note that the sum and difference of convex sets are convex and sois the intersection of any class of convex sets. However, the unionof two convex sets may not be convex.


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Convex functions

Convex functions

We say f : X 7→ R ∪ +∞ is convex if, for any x, y ∈ X andλ ∈ [0, 1],

f(λx+ (1− λ)y) ≤ λf(x) + (1− λ)f(y).


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Nonemptyness of subdifferential

The most useful property of a convex function related to Lagrangemultipliers is

Nonemptyness of subdifferential

Let f : X 7→ R ∪ +∞ be a convex function. Then for anyx ∈ int dom f ,

∂f(x) = ∅.

Prove and strengthening this result will be one of the focuses inthe next lecture.


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Epigraph characterizations

Epigraph characterizations

Function f : X 7→ R ∪ +∞ is convex iff epi f is a convex set inX × R.


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Subdifferential characterizations

Subdifferential characterizations

Function f : X 7→ R ∪ +∞ is convex iff, for anyx∗ ∈ ∂f(x), y∗ ∈ ∂f(y),

⟨y∗ − x∗, y − x⟩ ≥ 0.


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Cyclical monotonicity

In fact, we have the stronger result:

Cyclical monotonicity

Function f : X 7→ R ∪ +∞ is convex iff, for any m pairsx∗i ∈ ∂f(xi), i = 1, 2, . . . ,m,

⟨x∗1, x2 − x1⟩+ ⟨x∗2, x3 − x2⟩+ . . .+ ⟨x∗m, x1 − xm⟩ ≤ 0.


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Cyclical monotonicity: Proof

Simply add the following inequalities:

f(x2)− f(x1) ≥ ⟨x∗1, x2 − x1⟩f(x3)− f(x2) ≥ ⟨x∗2, x3 − x2⟩

. . . . . .

f(x1)− f(xm) ≥ ⟨x∗m, x1 − xm⟩.


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Corollary

Derivative characterizations

Function f : R 7→ R ∪ +∞ is convex iff f ′ is increasing orf ′′ ≥ 0.


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Simple operations

Operations

Suppose functions f, g : X 7→ R ∪ +∞ are convex andh : R → R is increasing then the following functions are convex:

• f + g.

• af, a > 0.

• h f .


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Supremum

Supremum

If fα : X 7→ R ∪ +∞ are convex then so is supα fα.

Key of the Proof:

epi supα

fα =∩

epi fα.


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Optimal value function

Optimal value function

Suppose that in problem v(y, z), f is convex, g is ≤K convex andh is affine and C is convex. Then the optimal value function v isconvex.


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Proof

Consider (yi, zi) ∈ dom v, i = 1, 2. ∀ε > 0, ∃xiε feasible to theconstraint of v(yi, zi) s.t. f(xiε) < v(yi, zi) + ε, i = 1, 2.Now for any λ ∈ [0, 1], we have

f(λx1ε + (1− λ)x2ε) ≤ λf(x1ε) + (1− λ)f(x2ε) (9)

< λv(y1, z1) + (1− λ)v(y2, z2) + ε.

Since λx1ε +(1−λ)x2ε is feasible for v(λ(y1, z1)+ (1−λ)(y2, z2)),v(λ(y1, z1) + (1− λ)(y2, z2)) ≤ f(λx1ε + (1− λ)x2ε). Combiningwith (9) and letting ε → 0 we arrive at

v(λ(y1, z1) + (1− λ)(y2, z2)) ≤ λv(y1, z1) + (1− λ)v(y2, z2),

that is to say v is convex.Peter Carr and Qiji Zhu CONVEX DUALITY IN MATH FINANCE

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Convex function related to convex sets

Let C be a closed convex set. Then the following functions areconvex:

• Distance function: dC . (View it as an optimal value function.)

• Support function: σC(x∗) = supx∈C⟨x∗, x⟩. (View it as

supremum of linear functions.)

• Indicator function: ιC .

• Gauge function: γC(x) = inft > 0 : x ∈ tC. (Immitate theproof of convexity of optimal value function.)


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Convex inf-convolution

Define the inf-convolution of f, g by

fg(x) = infy[f(x− y) + g(y)]

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Convexity of inf-convolution

If f, g are convex then so is fg.

Proof: fg(x) = inff(u) + r : g(y)− r ≤ 0, u+ y = x.


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Convexification

Often one need to generate a related convex function from one ora family of functions not necessarily convex. The following areseveral common methods suited for different applications:

• Integration of an increasing function.

• Convexification: Let f be an arbitrary function definef∗∗ = supg : g ≤ f and g convex.

• Regularized inf: Even if fα are all convex its inf is notnecessarily convex. But the regularized inf below is alwaysconvex:

ˆinfαfα = supg : g ≤ fα and g convex.


convex duality in math finance - homepages at wmuhomepages.wmich.edu/~zhu/mf/lecture1a.pdf ·...

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